Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts​ (a) through​ (c) below. TInterval ​(13.046,22.15) x overbarequals17.598 Sxequals16.01712719 nequals50 a. Express the confidence interval in the format that uses the​ "less than" symbol. Given that the original listed data use one decimal​ place, round the confidence interval limits accordingly. 13.05 Mbpsless thanmuless than 22.15 Mbps ​(Round to two decimal places as​ needed.) b. Identify the best point estimate of mu and the margin of error.

Answer:Point estimation of the mean = 17.6Error = 4.695% CI: 13.0 ≤ μ ≤ 22.2The distribution can be approximated to a normal because the sample size n=50 is bigger than 30.Step-by-step explanation:We have this information:- Sample mean:[tex]\bar{X}=17.598[/tex]- Sample standard deviation[tex]s=16.01712719[/tex]- Sample size[tex]n=50[/tex]For now on, we round to one decimal place.The best estimation for the population mean is the sample mean[tex]\mu=\bar{X}=17.6[/tex]The margin of error is equal to the t-value multiplied by the sample standard deviation and divided by the sample size.The t value depends on the degrees of freedom and the width of the confidence interval. In this case it is a 95% CI and the degrees of freedom are 49. For this conditions, the t-value is t=2.01.Then, the margin of error is[tex]E=t_{49}*\sigma/\sqrt{n}=2.01*16.0/\sqrt{50}=4.553[/tex]Then, the confidence interval can be constructed as:[tex]\bar{X}-t\sigma/\sqrt{n}\leq\mu\leq\bar{X}-t\sigma/\sqrt{n}\\\\17.6-4.6\leq \mu \leq 17.6+4.6\\\\ 13.0\leq \mu \leq 22.2[/tex]